Not sure how to approach this problem:
Let A be a commutative ring, I,J ideals of A such that A/I is a Noetherian ring. Show that J/(I ∩ J) is Noetherian as a A-module
So far I only concluded that if J were noetherian then A/(I ∩ J) would be too as isomorphic to product of noetherian.
From the isomorphism theorems, we know that $$\frac{I+J}{I} \cong \frac{J}{I\cap J}.$$ So $\frac{J}{I\cap J}$ is isomorphic as an $A/I$-module (and as an $A$-module) to $\frac{I+J}{I}$.
But $\frac{I+J}{I}$ is a submodule of $A/I$, which is Noetherian as an $A/I$-module. Submodules of Noetherian modules are Noetherian, so $\frac{I+J}{I}$ is a Noetherian $A/I$ module.
The $A$-module structure is the same as the $A/I$-module structure (same submodules), so $\frac{I+J}{I}$ is also a Noetherian $A$-module. Thus, $\frac{J}{I\cap J}$ is $A$-isomorphic to a Noetherian $A$-module, hence is a Noetherian $A$-module.